This is the answer on a): (Btw I didn't know how to translate one thing to english.. I just translated it to "digit sum" the digit sum of 123 is 1+2+3 = 6.
The difference between a positive integer and its digit sum is always divisible
with 9. We illustrate the proof idea for a three digit number 100a + 10b + c, where a, b
and c are digits. The sum is a + b + c, and the difference thus 99a + 9b =
9 * (11a + b). In particular, this difference is divisible by 3, and the familiar
rule that says that a number is divisible by 3 if and only if the digits are
it, and the corresponding rule for divisibility by 9, following light of this.
Let m = (1 +2 +3) + (4 +5 +6) + · · · + (4018 4019 4020) + (4021 4022),
where we have grouped the numbers three and three consecutive numbers with two digits to spare
eventually. The sum of three consecutive integers is always divisible by 3, and
is the sum of two consecutive integers is not divisible by 3, so
amount of each parenthesis is divisible by 3 - thus m
Let t be cross sum of n Then m - t divisible by 3, because m - t =
(1 + 2 + · · · + 4021 + 4022) - (1 + 2 + · · · + 4 + 0 + 2 + 1 + 4 + 0 + 2 + 2) =
(1-1) + (2-2) + · · · + (4021-4-0-2-1) + (4022-4-0-2-2) - each joint
in parentheses is the difference between a number and its digit sum, thus divisible
with 3 Since m and m - t is divisible by 3, is also not there - and thus also n
Alternatively, we can take account of cross sum more directly. The figures
0, 1,. . . , 999 each digit occurs 300 times (if we fill in with 0 in
the start of each number so all the three digit) so that the sum of all digits of
All these numbers are divisible by 3 - let's say that the sum is 3a. The sum of all
digits 0, 1,. . . , 3999 is thus 4 · 3a + 1000 · (1 + 2 + 3) = 3 · (4a + 2000)
thus also divisible by 3 The sum of all digits in 4000, 4001,. . . , 4022 is
23 · 4 + (10 · 1 + 3 · 2) + (2 · (1 + 2 + ... 9) + 1 + 2) = 102 + 3 · 2 + 2 · 45 + 3,
which is also divisible by 3 The sum of n - sum of all digits 1,
2,. . . , 4022 - is therefore divisible by 3, hence n
My english is not that good so I used google translate to help me, if something doesn't make sense please tell me. (I don't really understand it not even in my main language, this is just the solution they put up on the olympiad site..
Just double chcked b) and it's correctly typed in the question and everything else. I am not really this much of a genius in maths, I don't understand how a kid in my class can solve this.. :(
SPOILER: ANSWER ON B)
The digit sum t of a number with k er less digits fulfill 0 < t <= 9k < 10k. The number n has less than 4 * 4022 < 10^5 digits, so n < 10^10^5, and a1 = n^2011 < (10^10^5)^10^4 = 10^10^9, and a1 has therefore less digits than 10^9 and a digit sum that's less than 10^10. So a2 <10^10, has less than 10 digits and a digit sum that's less or equal to 81. So a3 <= 81 has the digit sum less than 18. (The only 2 digits number with such a huge digit sum is 99). So a4 <18. But a number is divisable by 9 if and only if the digit sum is divisable by 9, and a1 =n^2011 is divisiable with 9 since n is divisible with 3 - therefore is also a2, a3 and a4 divisible with 9. So 0 < a4 < 18 and a4 is divisible by 9, so a4 =9.
Note: If you see x <= 3 this just means that x is larger or equal to 3 (dunno how to type it properly in here)
Dunno is this makes any sense, makes zero sense to me (these tasks...)
). But practicing and reading the solutions gives you a general idea of what kind of solutions are expected.
